Question

Let a sequence be defined by $a_1=0$ and $a_{n+1}=a_n+4n+3$ for all $n\ge 1\left(nϵN\right)$.

If the value of $\underset{n\to \infty }{lim}\frac{\sqrt{a_n}+\sqrt{a_{9n}}+\sqrt{a_{9^{2}n}}+......+\sqrt{a_{9^{10}n}}}{\sqrt{a_n}+\sqrt{a_{3n}}+\sqrt{a_{3^{2}n}}+......+\sqrt{a_{3^{10}n}}}=L,$ then $\left[\frac{4L}{3^{11}}\right]$is
$\left[\right]$ denotes greatest integer function.

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is D $1$

As $a_{n+1}=a_n+4n+3$

$a_2=a_1+4.1+3=7=(2-1)(2.2+3)$

$a_3=a_2+4.2+3=18=(3-1)(3.2+3)$

$a_n=(n-1)(2n+3)$
$L={lim}_{n\to \infty }\frac{\sqrt{a_n}+\sqrt{a_{9n}}+\sqrt{a_{9^{2}n}}+......+\sqrt{a_{9^{10}n}}}{\sqrt{a_n}+\sqrt{a_{3n}}+\sqrt{a_{3^{2}n}}+......+\sqrt{a_{3^{10}n}}}$

$={lim}_{n\to \infty }\frac{\sqrt{(n-1)(2n+3)}+\sqrt{(9n-1)(2.9n+3)}+\sqrt{(9^{2}n-1)({2.9}^{2}n+3)}+......+\sqrt{(9^{10}n-1)({2.9}^{10}n+3)}}{\sqrt{(n-1)(2n+3)}+\sqrt{(3n-1)(2.3n+3)}+\sqrt{(3^{2}n-1)({2.3}^{2}n+3)}+......+\sqrt{(3^{10}n-1)({2.3}^{10}n+3)}}$

$={lim}_{n\to \infty }\frac{\sqrt{(1-\frac{1}{n})(2+\frac{3}{n})}+\sqrt{(9-\frac{1}{n})(2.9+\frac{3}{n})}+\sqrt{(9^2-\frac{1}{n})({2.9}^2+\frac{3}{n})}+......+\sqrt{(9^{10}-\frac{1}{n})({2.9}^{10}+\frac{3}{n})}}{\sqrt{(1-\frac{1}{n})(2+\frac{3}{n})}+\sqrt{(3-\frac{1}{n})(2.3+\frac{3}{n})}+\sqrt{(3^2-\frac{1}{n})({2.3}^2+\frac{3}{n})}+......+\sqrt{(3^{10}-\frac{1}{n})({2.3}^{10}+\frac{3}{n})}}$

$=\frac{\sqrt{\left(1\right)\left(2\right)}+\sqrt{\left(9\right)\left(2.9\right)}+\sqrt{\left(9^2\right)\left({2.9}^2\right)}+......+\sqrt{\left(9^{10}\right)\left({2.9}^{10}\right)}}{\sqrt{\left(1\right)\left(2\right)}+\sqrt{\left(3\right)\left(2.3\right)}+\sqrt{\left(3^2\right)\left({2.3}^2\right)}+......+\sqrt{\left(3^{10}\right)\left({2.3}^{10}\right)}}$

$=\frac{\sqrt{\left(1\right)\left(2\right)}\frac{1-9^{10}}{1-9}}{\sqrt{\left(1\right)\left(2\right)}\frac{1-3^{10}}{1-3}}=\frac{1-9^{10}}{4(1-3^{10})}$

Hence $\left[\frac{4L}{3^{11}}\right]=1$

Hence option ${}^{\prime }{D}^{\prime }$ is the answer.